A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using the equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second y=-16x^2+235x+133

Step-by-step explanation:

Given tq equation of the height reached by the rocket as y=-16x^2+235x+133

The velocity of the object at its maximum height is zero

v = dy/dx = 0(at max height)

v =-32x +235

0 =-32x+235

x = 235/32

x = 7.34

Hence the rocket will reach its maximum height after 7.34seconds

it has to have an angle of 90 degrees